If the lines frac{x - 1}{2} = frac{y + 1}{3} | vedclass

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(B) Let any point on the first line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = \lambda$ be $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$.Le

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$\frac{9}{2}$

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(B) Let any point on the first line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = \lambda$ be $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$.Let any point on the second line $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} = \mu$ be $(\mu + 3, 2\mu + k, \mu)$.The lines intersect if there exist $\lambda$ and $\mu$ such that the coordinates are equal:$2\lambda + 1 = \mu + 3 \implies 2\lambda - \mu = 2$ $(i)$$3\lambda - 1 = 2\mu + k \implies 3\lambda - 2\mu = k + 1$ (ii)$4\lambda + 1 = \mu \implies 4\lambda - \mu = -1$ (iii)Subtracting $(i)$ from (iii):$(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2$$2\lambda = -3 \implies \lambda = -\frac{3}{2}$.Substitute $\lambda = -\frac{3}{2}$ into (iii):$4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$.Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into (ii):$3(-\frac{3}{2}) - 2(-5) = k + 1$$-\frac{9}{2} + 10 = k + 1$$k = -\frac{9}{2} + 9 = \frac{9}{2}$.

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect,then $k =$

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